Atobacco plant is heterozygous for the genes for long stems (ss), large leaves (ll), and fibrous roots (rr). it is crossed with another tobacco plant with genotype ss ll rr. how many of the 80 offspring of this mating will be heterozygous for all three genes? these genes are not linked. hint: create three separate punnett squares and use the product rule (section 10.4c).

Firstly, you may start by doing the Punnett squares as suggested, one per each characteristic. Attached is the three Punnet squares, made by me, as they should look in the end.

For each of the Punnet squares, one characteristic each, we find that half of the offspring will be heterozygous for that characteristic.
Therefore, according to the product rule, you should now multiply the quantity of heterozygous for each characteristic and then obtain the total part of the offspring that's fully heterozygous.

Ss×Ll×Rr=SsLlRr

Total offspring×=10
There is 10 plants in the total offspring that are heterozygous for all the considered three genes.